I tried to prove example 3.4 from the book Morita Equivalence and Continuous-Trace C$^*$-Algebras by Iain Raeburn and Dana P. Williams, but I get uneasy with notations and ideas. Let me restate my assumption about this example.
Suppose that $T$ is locally compact Hausdorff space and that $H$ is a complex Hilbert space. Then $$X:=C_0(T,H)=\left\{ x:T\rightarrow H: x \:continuous,t\rightarrow \left\| x(t)\right\|\in C_0(T)\right\}$$ complex vector space and $$A:=C_0(T,K(H))=\left\{ f:T\rightarrow K(H): f \:continuous,t\mapsto \left\| x(t)\right\|\in C_0(T) \right\}$$ C*-algebra. Then $C_0(T,H)$ is left Hilbert-$C_0(T,K(H))$ module under this operation: $$f\bullet x(t)=f(t)(x(t))$$ $$_A \left\langle x,y \right\rangle (t)=x(t) \otimes \bar{y(t)} $$ for $x,y\in C_0(T,H)$ and $f\in C_0(T,K(H))$.
My questions are:
- Is $\otimes$ here related to tensor product? What kind of tensor product?
- Is $t\rightarrow \left\| x(t)\right\|\in C_0(T)$ and $t\mapsto \left\| x(t)\right\|\in C_0(T)$ different?
- Is norm for $_A\left\| x(t) \right\|$ a sup-norm?
For two vectors $x,y\in H$, the notation $x\otimes y$ is frequently used to denote the rank-one operator $z\longmapsto \langle z,y\rangle x$. It is related to tensor products, but not in the obvious way: it is seeing $x\otimes y$ as an element of $H\otimes H^*$ (most often, no one pays attention to this and just uses the notation).
I think it is just a typo, it should have been $\longmapsto $ in both cases.
I'm not sure what you mean by $_A\|x(t)\|$. The elements of the Hilbert module are functions $T\to H$. And $_A\langle x,y\rangle$ is a function $T\to K(H)$. In a Hilbert module you often construct a norm by taking the square root of the norm in $A$, i.e. $$ \|x\|:=\|_A\langle x,x\rangle\|_A^{1/2}. $$ In this case $A$ is a function algebra, so its norm is a sup norm. So $$ \|x\|=\sup\{|_A\langle x,x\rangle (t)|^{1/2}:\ t\in T\} $$